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mojhsa [17]
2 years ago
8

Mitosis is a process by which 1 cell divides and becomes 2 cells, those 2 cells divide and become 4 cells, and so on. Write an e

quation for the nth term of the geometric sequence giving the numbers of cells at each stage of this process, where a1=1. Then find a10.
Mathematics
1 answer:
uysha [10]2 years ago
4 0

Answer:

An equation for the nth term is aₙ = 2ⁿ⁻¹

and a₁₀ = 512

Step-by-step explanation:

From the question, 1 cell divides and becomes 2 cells, those 2 cells divide and become 4 cells, and so on

∴The sequence will be

1,2,4, ..., n

From the formula

aₙ = a₁rⁿ⁻¹

Where aₙ is the nth term

a₁ is the first term

n is the number of terms

and r is the common ratio

From the sequence,

a₁ = 1,

r = a₂/a₁ = 2/1 = 2

∴ The nth term (aₙ) will be

aₙ = (1)(2)ⁿ⁻¹

aₙ = 2ⁿ⁻¹

Hence, an equation for the nth term is aₙ = 2ⁿ⁻¹

To find a₁₀, substitute n= 10 into the equation

aₙ = 2ⁿ⁻¹

Then

a₁₀ = 2¹⁰ ⁻ ¹

a₁₀ = 2⁹

a₁₀ = 512

∴ a₁₀ = 512

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Given that,

The mass of a patient = 70 kg

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The patient donates 470mL of blood.

We know that,

1 pint = 568 mL

8 pints = 4544 mL

Required fraction,

\dfrac{470}{4544}=0.1\\\\=\dfrac{1}{10}

So, the required fraction is approximately \dfrac{1}{10}.

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2 years ago
The normal distribution An automobile battery manufacturer offers a 31/54 warranty on its batteries. The first number in the war
seraphim [82]

Answer:

1) if the manufacturer's assumptions are correct, it would reed to replace 0.62% of its batteries free.

2) a standard deviation of 6.0843 results in a 1.07% replacement rate

3) using the revised standard deviation for battery life, 91.9% of the manufacturer's batteries don't get free replacement but qualifies for the prorated credit

Step-by-step explanation:

based on the given data;

x will represent the random variable such that the lifetime of its auto batteries, is normally distributed with a mean of 45 months and a standard deviation of 5.6 months

so

x → N( U = 45, ∝ = 5.6)

Under the warranty, if a battery fails within 31 months of purchase, the manufacturer replaces the battery at no charges to the consumer.

if the battery fails after 31 months but within 54 months, the manufacturer provides a prostrated credit towards the purchase of anew battery

1) If the manufacturer's assumptions are correct,

p(x < 3) = p( [x-u / ∝ ] < [ 31-45 / 5.6] )

= p( z < -2.5 )

using the standard normal table,

value of z = 0.0062 ≈ 0.62%

so if the manufacturer's assumptions are correct, it would reed to replace 0.62% of its batteries free.

2)

The company finds that it is replacing 1.07% of its batteries free of charge. It suspects that its assumption standard deviation of the life of its batteries is incorrect, so a standard deviation of ? results in a 1.07%

so lets say;

p ( x < 31 ) = ( 1.07%) = 0.0107

p ( [x-u / ∝ ] < [ 31-45 / ∝] ) = 0.0107

now from the standard table

-2.301 is 1.07%

so

( 31 - 45 / ∝ ) = -2.301

-14 / ∝ = -2.301

∝ = -14 / - 2.301

∝ = 6.0843

therefore a standard deviation of 6.0843 results in a 1.07% replacement rate

3)

Using the revised standard deviation for battery life, what percentage of the manufacturer's batteries don't free replacement but do qualify for the prorated credit?

p( 31 < x < 54 ) = p ( [31 - u / ∝ ] < [ x-u / ∝]  < [ 54 - 45 / ∝] )

= p ( [31 - 45 / 6.0843 ] < [ x-u / ∝]  < [ 54 - 45 / 6.0843] )

= p ( -2.301 < z < 1.4792 )

= p(Z < 1.5) - p(Z < -2.3)

= 0.9393 - 0.0108

= 0.919 ≈ 91.9%

therefore using the revised standard deviation for battery life, 91.9% of the manufacturer's batteries don't get free replacement but qualifies for the prorated credit

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2 years ago
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